Cr atoms undergo a reduction reaction (decrease in oxidation number) from +6 to +3
Further explanation
Reduction oxidation reaction or abbreviated Redox is a chemical reaction in which there are changes in oxidation numbers
The general formula for determining oxidation numbers:
- 1. Single element atomic oxidation = 0. For example Ar, Mg, Cu, Fe, N₂, O₂, etc. = 0, Group IA (Li, Na, K, Rb, Cs, and Fr)= +1 ,Group IIA (Be, Mg, Ca, Sr and Ba)= +2
- H atoms in compounds = +1, except metal hydride compounds (Hydrogens that bind to groups IA or IIA) oxidation numbers H = -1, for example LiH, MgH₂, etc.
- 2. O atoms in compounds = -2, except OF2 = + 2 and at peroxide (Na₂O₂, BaO₂) = -1 and superoxide, for example KO₂ = -1/2.
- 3. The oxidation number in the uncharged compound = 0,
On reaction :
16H⁺ + 2Cr₂O₇²⁻ + C₂H₅OH → 4Cr³⁺+ 11H₂O + 2CO₂
Oxidation number at C₂H₅OH
2. C + 6. H + O = 0
2C + 6. + 1 + (-2) = 0
2C + 6 -2 = 0
2C + 4 = 0
2C = -4
C = -2
Oxidation number at CO₂
C + 2.O = 0
C + 2.-2 = 0
C = 4
The oxidation number C in C₂H₅OH is -2 and the oxidation number C in CO₂ is +4, the change in oxidation number C from -2 to +4 is 6, so the C atom experiences an oxidation reaction (increase in oxidation number)
Oxidation number at Cr₂O₇²⁻
2. Cr + 7. O = -2
2. Cr + 7.-2 = -2
2Cr -14 = -2
2Cr = 12
Cr = 6
Cr³⁺ = 3
The oxidation number Cr in Cr₂O₇²⁻ is +6 and the oxidation number Cr in Cr³⁺ is +3, the change in Cr oxidation number from +6 to +3 is -3, so that the Cr atom experiences a reduction reaction (decreasing the oxidation number)
While O and H atoms do not experience changes in oxidation numbers (O = -2 and H = +1)
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Keywords: reduction, oxidation
